Question : If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to:
Option 1: 18
Option 2: 12
Option 3: 15
Option 4: 16
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Correct Answer: 18
Solution :
If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to:
$(x+\frac{1}{x})^2 = x^2+2+ (\frac{1}{x})^2$
⇒ $(x+\frac{1}{x})^2 = 2+7$
⇒ $(x+\frac{1}{x})^2 = 9$
⇒ $(x+\frac{1}{x}) = 3$
⇒ $(x+\frac{1}{x})^3 = 3^3$
⇒ $x^3 + (\frac{1}{x})^3+3x\frac{1}{x}(x+\frac{1}{x}) = 27$
⇒ $x^3 + (\frac{1}{x})^3+3(3) = 27$
⇒ $x^3 + (\frac{1}{x})^3 = 27-9$
⇒ $x^3 + (\frac{1}{x})^3 = 18$
Hence, the correct answer is 18.
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